The lifetime of certain electronic components is a random variable with the expectation of 5000 hours and a standard deviation of 100 hours. What is the probability that the average lifetime of 400 components is less than 5012 hours?
When I calculated this answer I was not sure how to include 400 items into what I know about standard deviation but assuming the probability works much like calculating consecutive coin flips (Ex: 1 heads = 1/2^1, 2 heads = 1/2^2, n heads = 1/2^n) I just used the standard deviation of .12 and then added 400 as the exponent to get a ridiculously small number which makes me feel very unconfident.
My assumption:
P(400 items <= 5012 hours) = I(.12)^400 = .5478^400 = 2.810678e-105
Hint: Let the individual lifetimes be $X_1, X_2, X_3,\dots, X_n$ where $n=400$. Then the average lifetime $Y$ is given by $$Y=\frac{1}{n}\left(X_1+X_2+X_3+\cdots+X_n\right).$$ The mean of $Y$ is just $\dfrac{1}{n}$ times the sum of the means of the $X_i$. Each $X_i$ has mean $5000$, so the mean of $Y$ is $5000$.
The variance of an independent sum is the sum of the variances, and the variance of $kW$ is $k^2$ times the variance of $W$. It follows after a while that the variance of $Y$ is $\dfrac{10000}{400}$, and therefore the standard deviation of $Y$ is $\dfrac{100}{20}$.
Finally and most importantly, by the Central Limit Theorem we can probably safely assume that $Y$ has nearly normal distribution. I expect you can handle the rest. What is the probability that a normal with mean $5000$ and standard deviation $5$ is less than $5012$?